Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A plane EM wave travelling in vacuum along z-direction is given by

$E=E_{0}\sin (kz-\omega t)\hat{i}\; and \; B=B_{0}\sin (kz-\omega t)\hat{j}$

a) evaluate $\oint E.dl$ over the rectangular loop 1234 shown in the figure

b) evaluate $\int B.ds$ over the surface bounded by loop 1234

c) use equation $\oint E.dl=\frac{-d\phi _{B}}{dt}$ to prove $\frac{E_{0}}{B_{0}}=c$

d) by using a similar process and the equation

$\oint B.dl=\mu _{0}I\epsilon _{0}\frac{-d\phi _{E}}{dt}$ prove that $c=\frac{1}{\sqrt{\mu _{0}\epsilon _{0}}}$

Answers (2)

a)

$\oint \overrightarrow{E}.\overrightarrow{dl}=E_{0}h\left [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t) \right ]$

b)

$\oint \overrightarrow{B}.\overrightarrow{ds}=-\frac{B_{0}h}{k}\left [ \cos (kz_{2}-\omega t)-\cos (kz_{1}-\omega t) \right ]$

c) Substituting the above equations in the following equation we get

$\oint E.dl=-\frac{d\phi _{B}}{dt}=-\frac{d}{dt}\oint B.ds$

$\\E_{0}h\left [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t) \right ]=\frac{B_0hw}{k} [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]\\$

$\frac{E_{0}}{B_{0}}=C$

d)

$\oint B.dl=\mu _{0}I\epsilon _{0}\frac{-d\phi _{E}}{dt}$

$\\B_0 h[\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]=\frac{E_0h}{kw} [\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]$

$\\B_0 h=\frac{E_0h}{kw}$

We get

$c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}$

Posted by

infoexpert23

View full answer

[\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question